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Review: Quiz 2

Review: Quiz 2. Session 2f of “ Basic Electricity ” A Fairfield University E-Course Powered by LearnLinc. Basic Electricity. Two Sections Electron Flow and Resistance 5 on-line sessions Lab Inductance and Capacitance 5 on-line sessions Lab Mastery Test, Part 1.

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Review: Quiz 2

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  1. Review: Quiz 2 Session 2f of “Basic Electricity”A Fairfield University E-CoursePowered by LearnLinc Basic Electricity

  2. Basic Electricity Two Sections • Electron Flow and Resistance • 5 on-line sessions • Lab • Inductance and Capacitance • 5 on-line sessions • Lab Mastery Test, Part 1 Basic Electricity

  3. Basic Electricity(Continued) • Text:“Electricity One-Seven,” Harry Mileaf, Prentice-Hall, 1996, ISBN 0‑13‑889585‑6 (Covers several Modules and more) • References: • “Digital Mini Test: Principles of Electricity Lessons One and Two,” SNET Home Study Coordinator, (203) 771-5400 • Electronics Tutorial (Thanks to Alex Pounds) • Electronics Tutorial (Thanks to Mark Sokos) • Basic Math Tutorial (Thanks to George Mason University) • Vector Math Tutorial (Thanks to California Polytec atatom.physics.calpoly.edu ) Basic Electricity

  4. Section 2: AC, Inductors and Capacitors • 0BJECTIVES:This section introduces AC voltage / current and additional circuit components (inductors, transformers and capacitors). Basic Electricity

  5. Section 2 Schedule: Session 2a – 03/27Vector Math – 04/01 Session 2b – 04/03(Fri. Q&A session)Session 2c – 04/08Session 2d – 04/10(lab - 04/13, Sat.) Session 2e – 04/15Quiz 2 (due 04/22) Session 2f – 04/22 Fri. Q&A – 04/26 Sat. – 04/27 AC & Sine WavesSine Waves, Magnitude, Phase and VectorsInductors and CircuitsTransformersCapacitorsMore CapacitorsReview (Discuss Quiz 2) Review: Mastery Test Part 1 Mastery Test Part 1 Text 3.1 – 3.41Text 4.1 – 4.24 Text 3.42 – 3.73Text 3.74 – 3.100Text 3.101 – 3.135Text 3.135 – 3.148Text Chapter 2 Text Chap. 2 and 3 Basic Electricity

  6. Capacitor 2 Session Review • AC & capacitors • Capacitive reactance - Xc = 1 / (2 *  * f * C) • | I | = | V | / Xc • Current “Leads” Voltage (ICE) • DC and AC: Superposition • Power Factor [cos()] • Physical Capacitors • Mica • Ceramic • Electrolytic (polarized) • Adjustable (tuned “resonance”) The Water Model Current Voltage: pressure at bottom Basic Electricity

  7. Quiz_2 Results • The class had a B average – Nice Job. • Most of you should find the Mastery Test Part 1 easy. • 2 or 3 of you need to correct minor deficiencies. Basic Electricity

  8. Quiz True/False • Inductors block DC. (Capacitors block DC) _F_ • A larger inductor allows less current (for the same V & f). _T__ • An inductor allows more current at a higher frequency. _F__(XL = 2fL, higher reactance  less current) • A capacitor allows more current at a higher frequency. _T__ • A larger capacitor allows less current (for the same V & f). _F__ • Capacitors block DC. _T__ • Current leads voltage in an inductor _F__ • Current leads voltage in an capacitor _T__ • A transformer can change the voltage to current ratio at all _T_F_frequencies. (Doesn’t work at DC, f = 0) • A transformer can be used to “match” impedances. _T__ Basic Electricity

  9. Quiz Multiple Choice • A “Short Circuit” : • Is a small one (zero resistance) • Allows high current to flow • Stops current from flowing (only when a fuse blows) • Has a high voltage drop (actually zero voltage drop) • Solder is used to: • Make the circuit pretty • Glue the components to the circuit board (you should provide some mechanical strength separately) • Provide reliable electrical connection(wire nuts used for power) • None of the above • What value of inductor would you put in parallel with a 12H inductor to get an inductance of 4H? • 4k • 6k(Inductors in parallel add as inverses) • 8k • 12k Basic Electricity

  10. Quiz Multiple Choice (cont.) • What value of capacitor would you put in parallel with a 2F capacitor to get an capacitance of 5 F? • 1 F • 2 F(Capacitors in parallel add) • 3 F • 4 F • A capacitor can be placed across the contacts of a switch to: • Make the switch work at a higher voltage. • Protect the contacts from arcing due to an inductive load. (inductors attempt to keep the current flowing when the switch is opened) • Protect the contacts from arcing due to an capacitive load. • Make the switch work at higher frequencies. Basic Electricity

  11. R1 R5 6k R2 8k 12k R3 R4 12Vdc 6k 4k Circuit Analysis • For the following circuit treat R4 as a “load resistor” and find the Thevenin equivalent circuit. Basic Electricity

  12. R1 R5 6k R2 8k 12k R3 R4 12Vdc 6k 4k Thevenin 1a. Now treat resistor R4 as the “load” resistor. Find the Thevenin equivalent circuit. We are trying to find a simple circuit with one battery and one resistor in series that will behave the same as this circuit when we re-attach R4. • There are two steps • Find the voltage when the load is not attached • Find the value of the resistance to place in series with that voltage • Both of these are easier if we first reduce the circuit (no R4) Basic Electricity

  13. R5 R1,2 R1 R5 8 k 4 k 6k R2 8k R3 R4 12Vdc 6 k 4 k 12k R3 R4 12Vdc 6k 4k Thevenin (Cont.) 1b. Now simplify the parallel resistors R1 and R2 and use a voltage divider to find the “open circuit” voltage (VR3) The “open circuit” voltage (no R4) is the Thevenin voltage. We now have a “Voltage Divider” VR3 = VTH = 12V*4k/(8k+4k+4k) = 12*4/16 = 12*1/4 = 3V Basic Electricity

  14. R5 R1,2 R1 R5 8 k 4 k 6k R2 8k R3 R4 12Vdc 6 k 4 k 12k R3 R4 12Vdc 6k 4k R5,1,2 12 k V1 R3 R4 12Vdc 6k 4 k Thevenin (Cont.) 1c. Now find the Thevenin resistance. Now we need to find the Thevenin resistance. There are two ways to do this. The first is to “short out” output terminals and find the current that flows through the shorting element. ISC = 12V/12k = 1 mA, and RTH = VTH / ISC = 3V / 1 mA = 3 k Short Circuit Basic Electricity

  15. R5 R1,2 R1 R5 8 k 4 k 6k R2 8k R3 R4 12Vdc 6 k 4 k 12k R3 R4 12Vdc 6k 4k R4 4k R5,1,2 12 k R3 Thevenin (Cont.) 1d. Another way to find the Thevenin resistance. The second method for finding the Thevenin resistance is to set the voltage to zero and find the equivalent resistance “looking in” from the output terminals. RTH = 12k || 4k so RTH = 3k (1/12 + 1/4 = 4/12 = 1/3) 12k Basic Electricity

  16. R1 Rth R5 6k 3 k R2 8k Vth R4 3 Vdc 12k R3 R4 6 k 12Vdc 6k 4k Thevenin (Cont.) 1e. Here is the Thevenin equivalent circuit. We now have the Thevenin equivalent circuit. Basic Electricity

  17. Transient Circuit • In the circuit below, sketch the voltage across the 1 k resistor if the switch is closed at t = 0. 2a. Find the Time Constant: = L/R = 2 mH / 1 k = 2*10-3 / 103 = 2*10-6 = 2 sec Basic Electricity

  18. Transient Circuitcontinued 2.b Now sketch the voltage. It goes from 0 at t=0 to 10 ma as t gets large. 6.3 v 0 2s 4s 6s 8s 10s Basic Electricity

  19. Section 2 Schedule: Session 2a – 03/27Vector Math – 04/01 Session 2b – 04/03(Fri. Q&A session)Session 2c – 04/08Session 2d – 04/10(lab - 04/13, Sat.) Session 2e – 04/15Quiz 2 (due 04/22) Session 2f – 04/22 Fri. Q&A – 04/26 Sat. – 04/27 AC & Sine WavesSine Waves, Magnitude, Phase and VectorsInductors and CircuitsTransformersCapacitorsMore CapacitorsReview (Discuss Quiz 2) Review: Mastery Test Part 1 Mastery Test Part 1 Text 3.1 – 3.41Text 4.1 – 4.24 Text 3.42 – 3.73Text 3.74 – 3.100Text 3.101 – 3.135Text 3.135 – 3.148Text Chapter 2 Text Chap. 2 and 3 Basic Electricity

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